poj 3660 Cow Contest （有向图的传递闭包）

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6033		Accepted: 3262

DescriptionN (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.The contest is conducted in several head-to-head rounds, each between two cows. If cowA has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤B ≤ N; A ≠ B), then cow A will always beat cowB.Farmer John is trying to rank the cows by skill level. Given a list the results ofM (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.Input* Line 1: Two space-separated integers: N and M* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer,A, is the winner) of a single round of competition: A and BOutput* Line 1: A single integer representing the number of cows whose ranks can be determined　Sample Input

5 5
4 3
4 2
3 2
1 2
2 5

Sample Output

2

题意：n头牛进行比赛，给出m个胜负关系，问最后能确定等级的牛共有多少头。

思路：传递闭包。将没有直接关系的点通过floyd来传递，累加该点的入度和出度，若某个点的入度+出度=n-1，则说明该点与其他所有点的关系都已确定。

AC代码：

#include <cstring>#include <string>#include <cstdio>#include <algorithm>#include <queue>#include <cmath>#include <vector>#include <cstdlib>#include <iostream>#define max2(a,b) ((a) > (b) ? (a) : (b))#define min2(a,b) ((a) < (b) ? (a) : (b))using namespace std;int main(){int n,m,u,v;int dp[105][105];int in[105],out[105];scanf("%d%d",&n,&m);memset(dp,-1,sizeof(dp));memset(in,0,sizeof(in));memset(out,0,sizeof(out));while(m--){scanf("%d%d",&u,&v);dp[u][v]=0;in[v]++;out[u]++;}for(int k=1;k<=n;k++)for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(dp[i][k]>-1&&dp[k][j]>-1&&dp[i][j]==-1){dp[i][j]=0;in[j]++;out[i]++;}int ans=0;for(int i=1;i<=n;i++)if(in[i]+out[i]==n-1)ans++;printf("%d\n",ans);return 0;}

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